By bounding a certain quantity defined on real numbers by $f(n)$ I derived the following inequality arising from an inductive argument.
$ (n-2)/2 + f(n-1) \leq f(n).$
A solution to the above inequality is obtained by taking $f(n) = n^2.$ Since I want the best possible upper bound for the quantity in question I am wondering
Is there a function $f:\mathbb{N} \mapsto \mathbb{N}$ such that there is a fixed $N$ such that for all $n > N$ we have $$(n-2)/2 + f(n-1) \leq f(n)$$ and $$\lim_{n \to \infty} \frac{f(n)}{n^2} = 0?$$
It feels that for some fixed $\epsilon > 0$ the function $f(n) = n^{1+\epsilon}$ won't work so I am wondering if there is any other clever way to construct such an $f$ or perhaps whether it is not possible to do so at all?
Assume that $f(n)\geq f(n-1)+\frac{n-2}{2}$ holds for $n\geq N$ Then for large $n$, applying the inequality: $$f(n)\geq f(n-1)+\frac{n-2}{2}\geq f(n-2)+\frac{n-3}{2}+\frac{n-2}{2}\geq \ldots \geq f(N)+\sum_{j=N+1}^{n}\frac{j-2}{2}=$$ $$ =f(N)+\frac{n^2}{4}-\frac{3 n}{4}-\frac{N^2}{4}+\frac{3 N}{4} $$ Which is of order $n^2$